I have to show the equivalence of this Let $f(x) \in F[x]$, and $K / F$ an extension which contains $R_f$, the set of all root of $f(x)$. show the equivalence for a subfield $D \leq K$ :
(a) $D$ is the least element of the set $\left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$.
(b) $D$ is the minimal element of the set $\left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$.
(c) $D=\bigcap\left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$.
(d) $D=F\left(R_f\right)$.
for a) implies b)
I take $H \in \left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$ such that $H\subset D$ but how $D$ is the least element $D\subset H$ and then $D=H$, so $D$ is the minimal element.
for b) implies c) We know that $\bigcap\left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}\subset H$ for all $H \in \left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$ in particular for $D$ and how $D$ is minimal then $D= \bigcap\left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$
for c) implies d) How $F \leq L$ for all $L \in \left\{E \leq K \mid F \leq E, R_f \subseteq E\right\}$ the for all $\ell \in F(R_f)$ then $\ell \in L$ so $\ell \in D$ and then $F(R_f)\subseteq D$ but im stuck in showing the other contention, and c implies a, any help?